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Write a program which reads a sequence A of n elements and an integer M, and outputs "yes" if you can make M by adding elements in A, otherwise "no". You can use an element only once.

You are given the sequence A and q questions where each question contains Mi.

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers (Mi) are given.

Output

For each question Mi, print yes or no.

Constraints

• n ≤ 20
• q ≤ 200
• 1 ≤ elements in A ≤ 2000
• 1 ≤ Mi ≤ 2000

Sample Input 1

51 5 7 10 2182 4 17 8 22 21 100 35

Sample Output 1

nonoyesyesyesyesnono

Notes

You can solve this problem by a Burte Force approach. Suppose solve(p, t) is a function which checkes whether you can make t by selecting elements after p-th element (inclusive). Then you can recursively call the following functions:

solve(0, M)

The recursive function has two choices: you selected p-th element and not. So, you can check solve(p+1, t-A[p]) and solve(p+1, t) in solve(p, t) to check the all combinations.

For example, the following figure shows that 8 can be made by A[0] + A[2].

代码如下：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;const int maxn=25;int n,q;int a[maxn];int flag;int x;void cho (int loc,int sum){    if(sum==x)    {        flag=1;        return;    }    if(loc>n)        return;    cho (loc+1,sum);    cho (loc+1,sum+a[loc]);}void input (){    scanf("%d",&n);    for (int i=1;i<=n;i++)        scanf("%d",&a[i]);    scanf("%d",&q);    while (q--)    {        scanf("%d",&x);        flag=0;        cho (1,0);        if(flag)            printf("yes\n");        else            printf("no\n");    }}int main(){    input();    return 0;}
      
     
    
   

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